= Solution
{c}
The <altitude of a triangle> through $A$ has <equation> $(\mathbf a-\mathbf d)\cdot(\mathbf b-\mathbf c)=0$, and the <altitude of a triangle> through $B$ has <equation> $(\mathbf b-\mathbf d)\cdot(\mathbf a-\mathbf c)=0$. Their <normal vectors> $\mathbf b-\mathbf c$ and $\mathbf a-\mathbf c$ are <linearly independent> because the <triangle> is nondegenerate. Consequently these two <lines> meet in exactly one <point> $D$.
Substitute its <position vector> $\mathbf d$ in the supplied identity. Its first term vanishes by the first <altitude of a triangle> <equation>, and its second term vanishes by the second <altitude of a triangle> <equation>. Hence
$$
(\mathbf a-\mathbf b)\cdot(\mathbf c-\mathbf d)=0.
$$
This is exactly the condition that $D$ lies on the <altitude of a triangle> through $C$. \b[The three <lines> therefore meet in a unique <point>, the <orthocenter>.] This argument includes <right triangles> and <triangles> whose <orthocenter> lies outside them.
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